In this post, we will proof that repeatedly applying any algorithm on a solved Rubixs Cube would cause it to eventually return back to its solved state.

What do I mean by this? Well, supposed your algorithm is to turn the top side once. After repeating this algorithm 4 times, you would return back to the solved state right? In the problem above, “any algorithm” would refer to literally ANY ALGORITHM. For instance, your algorithm might be: R’ D D R D R’ D’ R, and repeating this algorithm a sufficient number of times on a solved cube would make it eventually return to its solved state.

This problem seems impossible right? At the very least, it probably involves some uni graph theory concept and stuff right? Before you close off (pls dont), the solution to this problem requires no prerequisites, just a grasp of logic.

Ok, here comes the solution:

To tackle this problem, I will be using this technique called “Proof by contradiction”.

For instance, if I want to proof that statement A is false, I assume otherwise; that A is true. Then I show that if A is true, absurd shit that doesn’t make sense would happen (ie. we would see a contradiction). Since A being true causes a contradiction, A must be false. Hence proved that A is false.

Got the general gist of this technique? Cause this would be fundamental in our proof.

Alright, the actual proof:

Let’s first assume that there exists an algorithm (Let’s call it “f”) that would not cycle. In other words, repeatedly applying “f” on a cube would cause it to always land in a state that it has not been before:

However, this would imply that a rubix cube has an INFINITE number of different states! This is a contradiction, since it is well known that a rubix cube has a finite number of states (Left as exercise of reader). Since there is a contradiction, we can safely conclude that after applying “f” sufficiently number of times, it would cause the cube to return to a state it was before.

We are not done however, as this conclusion results in 2 possible cases:

Case 1 is where it returns to the solve state (which we want), and Case 2 is where it returns not to the solved state but to some random state within the cycle (which we want to disprove).

To disprove case 2, we employ prove by contradiction again:

Let’s define “f^{-1}” as the inverse algorithm of “f”. In other words:

Basically, “f^{-1}” is an algorithm that REVERSES what “f” does.

So now, what happens if we replace all “f” with “f^{-1}” in the picture for Case 2? Well, it becomes this:

Now we will show how absurd this would be. Notice the box for “State n”. Notice how this diagram implies that applying the algorithm “f^{-1}” on state n would result in 2 possible states (State n-1 and State k). This is absurd because we know that applying an algorithm on a rubix cube will result in only 1 possible state (ie. each state should only point to 1 other state). Hence we have reached a conclusion that Case 2 is IMPOSSIBLE. Which leaves us with Case 1 as the only case that is possible.

Hence, proved, that repeatedly applying any algorithm on a solved Rubixs Cube would cause it to eventually return back to its solved state.

An update to the previous post: I have played with the simulator and recorded some of the animations

The dummy variable is basically an instruction of how to sum. For the example on the left, “n” is the dummy variable.

“a” is the starting point

“b” is the end point

and “f(n)” is the function you want to sum over.

What does this all mean?

Well, basically:

That’s it! Notice how “a” is the starting point and “b” is the end point?

Let’s take a look at some examples:

For the third example, note that “x” is the dummy variable and not y.

Now, what happens if you are walking along a beautiful path on a bright sunny day when suddenly a burly man pulls out a butterfly knife with a pearl-embedded, elephant tusk handle and put it against your throat. Adrenaline rushes through your veins as your limbs go numb with fear while your eyes watch the empty space ahead, futile in making sense of your awaiting doom. With a deep croak, your adversary spat

“We meet again. Now let’s talk.”

He pulls out a slip of paper. The paper cried as it flapped against the wind.

“Take a look. Take a close look at this. Explain.”

The slip of paper makes it’s way to inches from your face. Close up, you piece together the various symbols on the paper and break out in cold sweat. On the paper, largely printed, you saw:

Now what the hell is that “infinity” symbol doing there? Julian your master failed to cover this in his lectures, and you know you are doomed, for you don’t know what it means.

You don’t want to end up in the situation above do you? So listen up!

The infinity symbol up there means that you sum to infinity. You don’t stop summing, you just add and add and add and add…….. You are adding an infinite number of terms.

You don’t stop!

You might be asking now, wouldn’t the sum diverge to infinity? In other words:

Well the answer is: Yes, most of the time. MOST of the time.

You see, for some sums, it doesn’t approach infinity. For instance, you can proof that

Try it on your calculator! Or… research on Infinite Geometric Progressions… or not.

When it comes to infinite sums, there can be some quite unexpected results. For instance:

And yes, the π is indeed the one we are familiar with ( π=3.1415…)

To learn more about it, you can research on the Basel Problem, or the Riemann Zeta Function.

There are other types of summations, such as cyclic sum, symmetric sum and sum-over-n-that-satisfies-some-conditions (I’m don’t know the actual name for this type of sum, hopefully somebody can help fill me up on that). Anyway, I’m not going to go through those sums cause I’m lazy. Let’s move on to ∏ (product):

∏ is essentially the same as ∑, just that instead of summing, you multiply!

Now let’s take a look at random examples:

For the second example, note that “n” is the dummy variable and not “x”.

Similar to infinite sums, there are also infinite products, where you multiply FOREVER! HAHAHA!

Now, what happens if you are walking along a beautiful path on a bright sunny day when suddenly a burly man pulls out a butterfly knife with a pearl-embedded, elephant tusk handle and put it against your throat. Adrenaline rushes through your veins as your limbs go numb with fear while your eyes watch the empty space ahead, futile in making sense of your awaiting doom. With a deep croak, your adversary spat

“We meet again. Now let’s talk.”

He pulls out a slip of paper. The paper cried as it flapped against the wind.

“Take a look. Take a close look at this. Tell me. Tell me the exact value.”

The slip of paper makes it’s way to inches from your face. Close up, you piece together the various symbols on the paper and break out in cold sweat. On the paper, largely printed, you saw:

You faint like a Victorian lady, for you do not know.

Well, for this case, I’ll probably let you die, because I’m too lazy to give you a solution to the above problem. (Also because I’m not sure (Also because I don’t know)). However, I can give you the answer though:

Basically, my point is, just like an infinite sum, an infinite product can also yield interesting results.

Challenge: Using the above result regarding the infinite product, proof that:

If the LHS and the RHS exists. Hence, proof that:

First part of the challenge question can be solved with Sec 4 syllabus while the second part involves A-level syllabus. However, the second part is easier as compared to the first.

I hope that now you know what ∏ and ∑ means! Just don’t go around doing stuff like this: